How do I find the derivative of ln(e^(4x)+3x)?

Jun 21, 2016

$\left(f \left(g \left(x\right)\right)\right) ' = \frac{4 {e}^{4 x} + 3}{{e}^{4 x} + 3 x}$

Explanation:

We can find the derivative of this function using chain rule that says:

$\textcolor{b l u e}{\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)}$

Let us decompose the given function into two functions $f \left(x\right)$ and $g \left(x\right)$ and find their derivatives as follows:

$g \left(x\right) = {e}^{4 x} + 3 x$
$f \left(x\right) = \ln \left(x\right)$

Let's find the derivative of $g \left(x\right)$
Knowing the derivative of exponential that says:
$\left({e}^{u \left(x\right)}\right) ' = \left(u \left(x\right)\right) ' \cdot {e}^{u \left(x\right)}$
So,
$\left({e}^{4 x}\right) ' = \left(4 x\right) ' \cdot {e}^{4 x} = 4 {e}^{4 x}$
Then ,
$\textcolor{b l u e}{g ' \left(x\right) = 4 {e}^{4 x} + 3}$

Now Lets find $f ' \left(x\right)$

$f ' \left(x\right) = \frac{1}{x}$
According to the property above we have to find $f ' \left(g \left(x\right)\right)$ so let's substitute $x$ by $g \left(x\right)$ in $f ' \left(x\right)$ we have:

$f ' \left(g \left(x\right)\right) = \frac{1}{g} \left(x\right)$
$\textcolor{b l u e}{f ' \left(g \left(x\right)\right) = \frac{1}{{e}^{4 x} + 3 x}}$
Therefore,
$\left(f \left(g \left(x\right)\right)\right) ' = \left(\frac{1}{{e}^{4 x} + 3 x}\right) \cdot \left(4 {e}^{4 x} + 3\right)$

$\textcolor{b l u e}{\left(f \left(g \left(x\right)\right)\right) ' = \frac{4 {e}^{4 x} + 3}{{e}^{4 x} + 3 x}}$